Question

examine this figure. which two pieces of information, if true, would help to prove that \\( \triangle lmp \cong \triangle nmp \\) by hl? select two options. \\( \square \\) point p is the midpoint of \\( \overline{mk} \\). \\( \square \\) line mk is the perpendicular bisector of \\( \overline{ln} \\). \\( \square \\) \\( \overline{ml} \cong \overline{mp} \\) \\( \square \\) \\( \overline{ml} \cong \overline{mn} \\) \\( \square \\) \\( \overline{pk} \cong \overline{pk} \\)

Explanation:

Step1: Recall HL (Hypotenuse - Leg) theorem

HL theorem states that if the hypotenuse and a leg of a right - triangle are congruent to the hypotenuse and a leg of another right - triangle, then the two right - triangles are congruent.

Step2: Analyze the triangles \(\triangle LMP\) and \(\triangle NMP\)

For \(\triangle LMP\) and \(\triangle NMP\) to be right - triangles, \(MK\) must be the perpendicular bisector of \(LN\) (so \(\angle LPM=\angle NPM = 90^{\circ}\)).
If \(MK\) is the perpendicular bisector of \(LN\), then \(LP = NP\) (legs of the right - triangles).
For the hypotenuse, if \(\overline{ML}\cong\overline{MN}\) (hypotenuses of the right - triangles \(\triangle LMP\) and \(\triangle NMP\))

Answer:

Line \(MK\) is the perpendicular bisector of \(\overline{LN}\); \(\overline{ML}\cong\overline{MN}\)